Question

A homogeneous, simply supported prismatic beam of width B, depth D and span L is subjected to a concentrated load of magnitude P. The load can be placed anywhere along the span of the beam. The maximum flexural stress developed in beam is

• A. $\frac{2}{3}\frac{PL}{B{D}^2}$
• B. $\frac{3}{4}\frac{PL}{B{D}^2}$
• C. $\frac{4}{3}\frac{PL}{B{D}^2}$
• D. $\frac{3}{2}\frac{PL}{B{D}^2}$

Answer 1

The correct option is D $\frac{3}{2}\frac{PL}{B{D}^2}$


Using flexure formula,

$\frac{\sigma }{y}=\frac{M}{I}$

$\sigma =$ Flexural stress
$y=$ Distance from Neutral Axis
$M=$ Bending moment at the section
$I=$ Moment of Inertia about Neutral axis
$\sigma =\frac{My}{I}$

For $\sigma$ to be maximum, M should be maximum and y should also be maximum,

$y_{max}=\frac{D}{2}$

Maximum value of

$M=\frac{PL}{4}\left(Whenloadisatthemidspan\right)$

$I=\frac{B{D}^3}{12}$
$\sigma =\frac{My}{I}=\frac{\frac{PL}{4}.\frac{D}{2}}{\frac{B{D}^3}{12}}=\frac{3}{2}\frac{PL}{B{D}^2}$